Simplify the following expression and state the condition under which the simplification is valid. You can assume that $y \neq 0$. $n = \dfrac{16y - 56}{6} \div \dfrac{14y^2 - 49y}{-7} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $n = \dfrac{16y - 56}{6} \times \dfrac{-7}{14y^2 - 49y} $ When multiplying fractions, we multiply the numerators and the denominators. $n = \dfrac{ (16y - 56) \times -7 } { 6 \times (14y^2 - 49y) } $ $ n = \dfrac {-7 \times 8(2y - 7)} {6 \times 7y(2y - 7)} $ $ n = \dfrac{-56(2y - 7)}{42y(2y - 7)} $ We can cancel the $2y - 7$ so long as $2y - 7 \neq 0$ Therefore $y \neq \dfrac{7}{2}$ $n = \dfrac{-56 \cancel{(2y - 7})}{42y \cancel{(2y - 7)}} = -\dfrac{56}{42y} = -\dfrac{4}{3y} $